SSC MTS Model Paper 2

Question : 28 of 150

Marks: +1, -0

In a right angle ∆ABC, B = 90° and AC is hypotenuse. The perpendicular BD is dropped on hypotenuse AC from right angle vertex B, then

${ AD } = { { AB } ^ { 2 } } /{ \ { AC } }$
${ CD } = { { B C} ^ { 2 } } /{ \ { AC } }$
${ 1 } /{ { BD } ^ { 2 } } = { 1 } /{ { AB } ^ { 2 } } + { 1 }/ { { BC } ^ { 2 } }$
${ BD } = { { AB } × { BC } } /{ { AC } }$

Solution:

(2)

In ∆ABC and ∆BCD
∴ ∆ABC ~ ∆BCD (by AA)

{ BC } ^ { 2 } = { AC } × { CD }

{ { AC } } /{ { BC } } = { { AB } }/ { { BD } } = { { BC } }/ { { CD } } ⇒ { CD } = { { BC } ^ { 2 } }/ { { AC } }

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